Shashank Shekhar

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# Rings and Rods | Leetcode Problem Solution Using Bit Manipulation

## Problem Statement

There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.

You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:

• The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B').
• The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9').

For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.

Return the number of rods that have all three colors of rings on them.

Sample Input
B0B6G0R6R0R6G9

Sample Output
1

Explanation

• The rod labeled 0 holds 3 rings with all colors: red, green, and blue.
• The rod labeled 6 holds 3 rings, but it only has red and blue.
• The rod labeled 9 holds only a green ring. Thus, the number of rods with all three colors is 1.

## Solution (C++)

``````class Solution {
public int countPoints(String rings) {
int count, n=rings.length();
int rods[]=new int[10];
Arrays.fill(rods, 0);
for(int i=0; i<n; i+=2) {
char color=rings.charAt(i);
int rod=rings.charAt(i+1)-48;
switch(color) {
case 'R':
rods[rod]=rods[rod] | 1;
break;
case 'G':
rods[rod]=rods[rod] | 2;
break;
case 'B':
rods[rod]=rods[rod] | 4;
break;
}
}
count=0;
for(int item: rods)
count+=item==7?1:0;
return count;
}
};
``````